Phy5646/AddAngularMomentumProb

Based on exercise 15.1.1. from Principles of Quantum Mechanics, 2nd ed. by Shankar:

Express $\ S^{2}$ as a matrix for two spin-1/2 particles in the direct product basis.

1.) First express $\ S^{2}$ in terms of $\ S_{1}^{2}$, $\ S_{2}^{2}$, $\ S_{1z}$, $\ S_{2z}$, $\ S_{1\pm }$ and $\ S_{2\pm }$: $\ S^{2}=({\vec {S_{1}}}+{\vec {S_{2}}})^{2}=S_{1}^{2}+S_{2}^{2}+2{\vec {S_{1}}}\cdot {\vec {S_{2}}}=S_{1}^{2}+S_{2}^{2}+2(S_{1x}S_{2x}+S_{1y}S_{2y}+S_{1z}S_{2z})=S_{1}^{2}+S_{2}^{2}+2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+}$

2.) Then act with this on direct product state $\ |m_{1}m_{2}\rangle$: $\ S^{2}|m_{1}m_{2}\rangle =(S_{1}^{2}+S_{2}^{2}+2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+})|m_{1}m_{2}\rangle$ $\ =S_{1}^{2}|m_{1}m_{2}\rangle +S_{2}^{2}|m_{1}m_{2}\rangle +2S_{1z}S_{2z}|m_{1}m_{2}\rangle +S_{1+}S_{2-}|m_{1}m_{2}\rangle +S_{1-}S_{2+}|m_{1}m_{2}\rangle$

$\ =\hbar ^{2}{\frac {1}{2}}\left({\frac {1}{2}}+1\right)|m_{1}m_{2}\rangle +\hbar ^{2}{\frac {1}{2}}\left({\frac {1}{2}}+1\right)|m_{1}m_{2}\rangle +2\hbar ^{2}m_{1}m_{2}|m_{1}m_{2}\rangle +\hbar ^{2}{\sqrt {{\frac {1}{2}}\left({\frac {1}{2}}+1\right)-m_{1}\left(m_{1}+1\right)}}{\sqrt {{\frac {1}{2}}\left({\frac {1}{2}}+1\right)-m_{2}\left(m_{1}-1\right)}}|m_{1}+1;m_{2}-1\rangle$ $\ +\hbar ^{2}{\sqrt {{\frac {1}{2}}\left({\frac {1}{2}}+1\right)-m_{1}\left(m_{1}-1\right)}}{\sqrt {{\frac {1}{2}}\left({\frac {1}{2}}+1\right)-m_{2}\left(m_{1}+1\right)}}|m_{1}-1;m_{2}+1\rangle$ $\ \Rightarrow S^{2}|m_{1}m_{2}\rangle =\hbar ^{2}\left(\left({\frac {3}{2}}+2m_{1}m_{2}\right)|m_{1}m_{2}\rangle +{\sqrt {\left({\frac {3}{4}}-m_{1}(m_{1}+1)\right)\left({\frac {3}{4}}-m_{2}(m_{2}-1)\right)}}|m_{1}+1;m_{2}-1\rangle +{\sqrt {\left({\frac {3}{4}}-m_{1}(m_{1}-1)\right)\left({\frac {3}{4}}-m_{2}(m_{2}+1)\right)}}|m_{1}-1;m_{2}+1\rangle \right)$

3.) Now acting on the left with $\ \langle m_{1}'m_{2}'|$:

$\ \langle m_{1}'m_{2}'|S^{2}|m_{1}m_{2}\rangle =\hbar ^{2}\left(\left({\frac {3}{2}}+2m_{1}m_{2}\right)\delta _{m_{1}'m_{1}}\delta _{m_{2}'m_{2}}+{\sqrt {\left({\frac {3}{4}}-m_{1}(m_{1}+1)\right)\left({\frac {3}{4}}-m_{2}(m_{2}-1)\right)}}\delta _{m_{1}'m_{1}+1}\delta _{m_{2}'m_{2}-1}+{\sqrt {\left({\frac {3}{4}}-m_{1}(m_{1}-1)\right)\left({\frac {3}{4}}-m_{2}(m_{2}+1)\right)}}\delta _{m_{1}'m_{1}-1}\delta _{m_{2}'m_{2}+1}\right)$

$\Rightarrow \langle m_{1}'m_{2}'|S^{2}|m_{1}m_{2}\rangle =\hbar ^{2}\left(\left({\frac {3}{2}}+2m_{1}m_{2}\right)\delta _{m_{1}'m_{1}}\delta _{m_{2}'m_{2}}+{\sqrt {\left({\frac {3}{4}}-m_{1}m_{1}'\right)\left({\frac {3}{4}}-m_{2}m_{2}'\right)}}\left(\delta _{m_{1}'m_{1}+1}\delta _{m_{2}'m_{2}-1}+\delta _{m_{1}'m_{1}-1}\delta _{m_{2}'m_{2}+1}\right)\right)$

3.) Now plugging in appropriate values of $\ m_{1},m_{2},m_{1}'$ and $\ m_{2}'$:

$\ \langle 1/2;1/2|S^{2}|1/2;1/2\rangle =\hbar ^{2}\left(\left({\frac {3}{2}}+2\cdot {\frac {1}{2}}\cdot {\frac {1}{2}}\right)+0\right)=2\hbar ^{2}$

$\ \langle -1/2;-1/2|S^{2}|-1/2;-1/2\rangle =\hbar ^{2}\left(\left({\frac {3}{2}}+2\cdot {\frac {-1}{2}}\cdot {\frac {-1}{2}}\right)+0\right)=2\hbar ^{2}$

$\langle 1/2;-1/2|S^{2}|1/2;-1/2\rangle =\left(\left({\frac {3}{2}}+2\cdot {\frac {1}{2}}\cdot {\frac {-1}{2}}\right)+0\right)=\hbar ^{2}$

$\langle -1/2;1/2|S^{2}|-1/2;1/2\rangle =\left(\left({\frac {3}{2}}+2\cdot {\frac {-1}{2}}\cdot {\frac {1}{2}}\right)+0\right)=\hbar ^{2}$

$\langle -1/2;1/2|S^{2}|1/2;1/2\rangle =\langle 1/2;1/2|S^{2}|-1/2;1/2\rangle =\left(0+0\right)=0$

$\langle 1/2;-1/2|S^{2}|1/2;1/2\rangle =\langle 1/2;1/2|S^{2}|1/2;-1/2\rangle =\left(0+0\right)=0$

$\langle -1/2;-1/2|S^{2}|1/2;1/2\rangle =\langle 1/2;1/2|S^{2}|-1/2;-1/2\rangle =\left(0+0\right)=0$

$\langle -1/2;1/2|S^{2}|1/2;-1/2\rangle =\langle 1/2;-1/2|S^{2}|-1/2;1/2\rangle =\left(0+{\sqrt {\left({\frac {3}{4}}-{\frac {1}{2}}\cdot {\frac {-1}{2}}\right)\left({\frac {3}{4}}-{\frac {-1}{2}}\cdot {\frac {1}{2}}\right)}}\left(0+1\right)\right)=\hbar ^{2}$

$\langle -1/2;-1/2|S^{2}|-1/2;1/2\rangle =\langle -1/2;1/2|S^{2}|-1/2;-1/2\rangle =\left(0+0\right)=0$

$\langle -1/2;-1/2|S^{2}|1/2;-1/2\rangle =\langle 1/2;-1/2|S^{2}|-1/2;-1/2\rangle =\left(0+0\right)=0$

All that must be done now is arranging the matrix elements in matrix form. The ++ state corresponds to the left (top) while the -- state is on the right (bottom):

$S^{2}={\hbar }^{2}{\begin{pmatrix}2&0&0&0\\0&1&1&0\\0&1&1&0\\0&0&0&2\\\end{pmatrix}}$

We can clearly see that all of the direct product states do not diagonalize $\ S^{2}$. A linear combination of the two problem states, +- and +-, should solve the problem however:

$\ S^{2}(\alpha |+-\rangle +\beta |-+\rangle )=\lambda (\alpha |+-\rangle +\beta |-+\rangle )$

$\ \Rightarrow (S_{1}^{2}+S_{2}^{2}+2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+})(\alpha |+-\rangle +\beta |-+\rangle )=\lambda (\alpha |+-\rangle +\beta |-+\rangle )$

$\ \Rightarrow \hbar ^{2}\alpha \left(\left({\frac {3}{4}}+{\frac {3}{4}}+2\left({\frac {1}{2}}\cdot {\frac {-1}{2}}\right)\right)|+-\rangle +0+{\sqrt {\left({\frac {3}{4}}+{\frac {1}{4}}\right)\left({\frac {3}{4}}+{\frac {1}{4}}\right)}}|-+\rangle \right)+\hbar ^{2}\beta \left(\left({\frac {3}{4}}+{\frac {3}{4}}+2\left({\frac {-1}{2}}\cdot {\frac {1}{2}}\right)\right)|-+\rangle +{\sqrt {\left({\frac {3}{4}}+{\frac {1}{4}}\right)\left({\frac {3}{4}}+{\frac {1}{4}}\right)}}|+-\rangle +0\right)=\lambda (\alpha |+-\rangle +\beta |-+\rangle )$

$\ \Rightarrow \hbar ^{2}\alpha \left(|+-\rangle +|-+\rangle \right)+\hbar ^{2}\beta \left(|-+\rangle +|+-\rangle )\right)=\lambda (\alpha |+-\rangle +\beta |-+\rangle )$

$\ \Rightarrow \hbar ^{2}\left(\alpha +\beta \right)\cdot \left(|+-\rangle +|-+\rangle \right)=\lambda (\alpha |+-\rangle +\beta |-+\rangle )$